Five times the sum of three consecutive integers is 150. what are the integers?

1 answer:

8 0

Alright! Let's start off with the basic from the info given:

150= 5(x+y+z)
First off, since you're looking for x y and z, let's divide both sides by 5:
30= x+y+z

Now let's change y and z written in terms of x:
y= x+1 because it's a consecutive integer
z=x+2 because it's 1 + the value for y

So let's set this up:
30= x + x+1 +x+2

Then simplify
30= 3x + 3

Since you're looking for x, let's subtract 3 then divide by 3
27= 3x
9=x

Now that you have the value of x, plug it into the equations above for y and z:
y=x+1
y= 9+1=10

z= x+2
z= 9+2 =11

Try it out!

(9+10+11) * 5 = 150
(30) * 5 = 150

150 = 150!

So you have x= 9, y= 10, and z=11

Hope this helps!

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Which expression is equivalent??? help!

dexar [7]

Answer:

The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

Step-by-step explanation:

Given:

$\sqrt[3]{256x^{10}y^{7} }$

Solution:

We will see first what is Cube rooting.

$\sqrt[3]{x^{3}} = x$

Law of Indices

$(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}$

Now, applying above property we get

$\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})$